12-30 Wannafly每日一题 Pretty Song

来源：互联网发布：矩阵满秩是什么意思编辑：程序博客网时间：2024/05/16 07:56

【题目链接】点击打开链接

【解题方法】

求一个字符串的所有字串的权值和，每个字串的权值为元音字母的个数比上字串的长度将字串转化为01串，那么区间[l,r]的字串的权值为（s[r]-s[l-1]）/（r-l+1），枚举长度k，则所有字串的权值和为Sigma(1/k *(s[k]-[s0] + s[k+1]-s[1]+...s[n]-s[n-k]) )       一式令sum[i]=s[0]+s[1]+s[2]+...+s[i]则一式转化为sum[n]-sum[k-1]-sum[n-k]

【AC代码】

#include <set>#include <map>#include <queue>#include <stack>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>using namespace std;typedef long long LL;const int maxn = 5e5+20;char s[maxn];int a[maxn];LL sum[maxn];int main(){    while(scanf("%s",s+1)!=EOF)    {        int len = strlen(s+1);        for(int i = 1; i <= len; i++){            if(s[i] == 'I' || s[i] == 'E' || s[i] == 'A' || s[i] == 'O' || s[i] == 'U' || s[i] == 'Y'){                a[i] = 1;            }            else{                a[i] = 0;            }        }        memset(sum, 0, sizeof(sum));        for(int i = 1; i <= len; i++){            a[i] += a[i-1];            sum[i] = sum[i-1] + a[i];        }        double ans = 0.0;        for(int k = 1; k <= len; k++)        {            ans = ans + (double)(sum[len] - sum[k - 1] - sum[len - k]) / k*1.0;        }        printf("%.8f\n", ans);    }    return 0;}

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